Order-complete chains have a very special property. Suppose that $X$ and $Y$ are chains and $X_0$ is a subset of $X,$ and that $f$ is an order-preserving function from $X_0$ to $Y\,.$ The problem is: Does there exist an isotone extension of $f$ whose domain is $X\,?$ Unless some restriction is made on $f,$ the answer is "no", for if $X$ is the set of all positive real numbers, $X_0$ is the subset consisting of all the numbers which are less then one, $Y= X_0$ and $f$ is the identity map, then it is easy to see that there is no isotone function.
(Assuming an extension $f^{-},$ what is $f^{-}(1)\,$?) But this example also indicates the nature of difficulty, for $X_0$ is a subset of $X$ which has an upper bound and $f[X_0]$ has no upper bound. If an isotone extension $f^-$ exists, then the image under $f^-$ of an upper bound for a set $A$ is surely an upper bound for $f[A]\,.$
The above excerpt is taken from General Topology by Kelley; here he discusses whether an isotone extension exists or not.
I didn't get the example.
Why, if $X_0 =\{x\in \mathbb R: x\lt 1\}$ and $Y= X_0,$ there is no isotone function?
How come $X_0$ has an upper bound and not $f[X_0]\,?$
What/Why is basically the "restriction" the author is talking about, necessary to be an isotone extension?
Clearly $X_0$ has an upper bound in $\Bbb R$, for example $1$. But in $X_0$ it has no upper bound, because that would correspond to a maximal element strictly below $1$ in the real numbers.
So how can we extend the identity function to remain monotone? It would have to map bounded sets to bounded sets, but it is impossible to bound $X_0$ in $X_0$, so you cannot map any upper bound of $X_0$ from $\Bbb R$ into $X_0$ by extending the identity function and preserving monotonicity.